Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x+y &= 7 \\ 4x-y &= 8\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $4x = y+8$ Divide both sides by $4$ to isolate $x$ $x = {\dfrac{1}{4}y + 2}$ Substitute this expression for $x$ in the first equation. $5({\dfrac{1}{4}y + 2}) + y = 7$ $\dfrac{5}{4}y + 10 + y = 7$ Simplify by combining terms, then solve for $y$ $\dfrac{9}{4}y + 10 = 7$ $\dfrac{9}{4}y = -3$ $y = -\dfrac{4}{3}$ Substitute $-\dfrac{4}{3}$ for $y$ in the top equation. $5x- \dfrac{4}{3} = 7$ $5x-\dfrac{4}{3} = 7$ $5x = \dfrac{25}{3}$ $x = \dfrac{5}{3}$ The solution is $\enspace x = \dfrac{5}{3}, \enspace y = -\dfrac{4}{3}$.